Question: Let $z_1,$ $z_2,$ $\dots,$ $z_{20}$ be the twenty (complex) roots of the equation
\[z^{20} - 4z^{19} + 9z^{18} - 16z^{17} + \dots + 441 = 0.\]Calculate $\cot \left( \sum_{k = 1}^{20} \operatorname{arccot} z_k \right).$  Note that the addition formula for cotangent is still valid when working with complex numbers.
Solution: We start with the addition formula for tangent:
\[\tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}.\]Then
\begin{align*}
\cot (a + b) &= \frac{1}{\tan (a + b)} \\
&= \frac{1 - \tan a \tan b}{\tan a + \tan b} \\
&= \frac{\frac{1}{\tan a \tan b} - 1}{\frac{1}{\tan a} + \frac{1}{\tan b}} \\
&= \frac{\cot a \cot b - 1}{\cot a + \cot b}.
\end{align*}Then
\begin{align*}
\cot (a + b + c) &= \cot ((a + b) + c) \\
&= \frac{\cot (a + b) \cot c - 1}{\cot (a + b) + \cot c} \\
&= \frac{\frac{\cot a \cot b - 1}{\cot a + \cot b} \cdot \cot c - 1}{\frac{\cot a \cot b - 1}{\cot a + \cot b} + \cot c} \\
&= \frac{\cot a \cot b \cot c - (\cot a + \cot b + \cot c)}{(\cot a \cot b + \cot a \cot c + \cot b \cot c) - 1}.
\end{align*}More generally, we can prove that
\[\cot (a_1 + a_2 + \dots + a_n) = \frac{s_n - s_{n - 2} + \dotsb}{s_{n - 1} - s_{n - 3} + \dotsb},\]where $s_k$ is the sum of the products of the $\cot a_i,$ taken $k$ at a time.  (In the numerator, the terms are $s_n,$ $s_{n - 2},$ $s_{n - 4},$ $s_{n - 6},$ $\dots,$ and the signs alternate.  The numerator ends at $s_0 = 1$ or $s_1,$ depending on whether $n$ is even or odd.  The terms in the denominator are similarly described.)

Let $a_i = \operatorname{arccot} z_i.$  Then
\[\cot (a_1 + a_2 + \dots + a_{20}) = \frac{s_{20} - s_{18} + \dots - s_2 + 1}{s_{19} - s_{17} + \dots + s_3 - s_1}.\]By Vieta's formulas, $s_1 = 2^2,$ $s_2 = 3^2,$ $s_3 = 4^2,$ $\dots,$ $s_{19} = 20^2,$ and $s_{20} = 21^2.$  Therefore,
\begin{align*}
\cot (a_1 + a_2 + \dots + a_{20}) &= \frac{s_{20} - s_{18} + \dots - s_2 + 1}{s_{19} - s_{17} + \dots + s_3 - s_1} \\
&= \frac{21^2 - 19^2 + 17^2 - 15^2 + \dots + 5^2 - 3^2 + 1}{20^2 - 18^2 + 16^2 - 14^2 + \dots + 4^2 - 2^2} \\
&= \frac{(21 - 19)(21 + 19) + (17 - 15)(17 + 15) + \dots + (5 - 3)(5 + 3) + 1}{(20 - 18)(20 + 18) + (16 - 14)(16 + 14) + \dots + (4 - 2)(4 + 2)} \\
&= \frac{2(21 + 19 + 17 + 15 + \dots + 5 + 3) + 1}{2(20 + 18 + 16 + 14 + \dots + 4 + 2)} \\
&= \boxed{\frac{241}{220}}.
\end{align*}